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Hint:For the above question we have been given a composite function to find its derivative and we know that to find the derivative of a composite function we will have to use the chain rule of derivative which states that the derivative of a composite function,

\[F\left( g\left( x \right) \right)\] is \[F'\left( g\left( x \right) \right)\times g'(x)\]

__Complete step-by-step answer:__

We have been asked to find the derivative of the function, \[\cos \left( \log x+{{e}^{x}} \right)\].

Since it is a composite function, so we will use the chain rule of derivative to find its derivatives and the chain rule states that the derivative of a function \[F\left( g\left( x \right) \right)\] is equal to \[F'\left( g\left( x \right) \right)\times g'(x)\].

So the differentiation of \[\cos \left( \log x+{{e}^{x}} \right)\] is shown as follows by applying the chain rule:

\[\dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log +{{e}^{x}} \right)\times \dfrac{d}{dx}\left( \log x+{{e}^{x}} \right)\]

Since we know that the derivative of cos x is –sin x.

\[\Rightarrow -\sin \left( \log x+{{e}^{x}} \right)\times \left[ \dfrac{d}{dx}\left( \log x \right)+\dfrac{d}{dx}\left( {{e}^{x}} \right) \right]\]

As we know the derivative of log x is \[\dfrac{1}{x}\] and the derivative of \[{{e}^{x}}\] is \[{{e}^{x}}\], we can write the above function as follows.

\[\Rightarrow \dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\]

Therefore, the required derivative of the given function is equal to \[-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\].

Note: Be careful while applying the chain rule to the composite function and also take care of the sign. Sometimes we take the derivative of cos x as sin x by mistake and we just forget the negative sign before sin x, so be careful at that time otherwise we will get incorrect answers.

\[F\left( g\left( x \right) \right)\] is \[F'\left( g\left( x \right) \right)\times g'(x)\]

We have been asked to find the derivative of the function, \[\cos \left( \log x+{{e}^{x}} \right)\].

Since it is a composite function, so we will use the chain rule of derivative to find its derivatives and the chain rule states that the derivative of a function \[F\left( g\left( x \right) \right)\] is equal to \[F'\left( g\left( x \right) \right)\times g'(x)\].

So the differentiation of \[\cos \left( \log x+{{e}^{x}} \right)\] is shown as follows by applying the chain rule:

\[\dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log +{{e}^{x}} \right)\times \dfrac{d}{dx}\left( \log x+{{e}^{x}} \right)\]

Since we know that the derivative of cos x is –sin x.

\[\Rightarrow -\sin \left( \log x+{{e}^{x}} \right)\times \left[ \dfrac{d}{dx}\left( \log x \right)+\dfrac{d}{dx}\left( {{e}^{x}} \right) \right]\]

As we know the derivative of log x is \[\dfrac{1}{x}\] and the derivative of \[{{e}^{x}}\] is \[{{e}^{x}}\], we can write the above function as follows.

\[\Rightarrow \dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\]

Therefore, the required derivative of the given function is equal to \[-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)\].

Note: Be careful while applying the chain rule to the composite function and also take care of the sign. Sometimes we take the derivative of cos x as sin x by mistake and we just forget the negative sign before sin x, so be careful at that time otherwise we will get incorrect answers.

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